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3 years ago

18. [Algebra - Expansion] Expand 3(x + 2)

2 answers:

7 0

First you you distribute the 3 (3 times x and 3 times 2)
3x+6
Then divide 3x on both sides (3x divided by 3x and 6)
Your answer should be x=2

alexira [117]3 years ago

4 0

Answer:

Step-by-step explanation:

3(x + 2) = 3*x + 3*2

= 3x + 6

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(-2h3n2)(-5h2n2)<br> How

IRINA_888 [86]

Answer: 9h x 8n^2

we rearrange this underneath the question (-5h-2h) (2nx3n)(2+2) or if 2^2 see explanation. It must therefore be 7h6n4

Step-by-step explanation:

if its ^2 )(power2) then the second reasonable answer is (7h-2h)(2n^2 x 2n^2)=9h x 4n^2 x4n^2 =9h 8n^2 or 17hn^2 but i leave as 9h 8n^2

You can expand only if multiplying 9hx4 for example but not combine the letters. The questions that ask you to mix are much different to this. but if you want to add them 9h+8n^2 =17hn^2

3 0

2 years ago

2(x + 1 ) - 3(x + 5) ≥ 0 please solve!!

djverab [1.8K]

Answer:

Step-by-step explanation:

2x + 2 -3x -15 > 0

-x-13 > 0

-x > 13

Answer: x < -13

8 0

2 years ago

Read 2 more answers

Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat

drek231 [11]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

Step-by-step explanation:

For this case we have the following probability distribution given:

X 0 1 2 3 4 5

P(X) 0.031 0.156 0.313 0.313 0.156 0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

We can verify that:

$\sum_{i=1}^n P(X_i) = 1$

And $P(X_i) \geq 0, \forall x_i$

So then we have a probability distribution

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

6 0

2 years ago

(I WILL MARK BRAINLY)

Mademuasel [1]

Answer:

Step-by-step explanation:

light blue, 4,6,8

5 0

2 years ago

WILL MARK BRAINLIEST!!!!!!!!! <br> what is 5/1 as a whole number

MariettaO [177]

Answer:

Step-by-step explanation:

5÷1

i m not sure but i like its ans

3 0

2 years ago